CHE302 Process Dynamics and Control Korea University 9-1
CHE302 LECTURE IX
FREQUENCY RESPONSES
Professor Dae Ryook Yang
Fall 2001
Dept. of Chemical and Biological Engineering Korea University
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DEFINITION OF FREQUENCY RESPONSE
• For linear system
– “The ultimate output response of a process for a sinusoidal input at a frequency will show amplitude change and phase shift at the same frequency depending on the process characteristics.”
– Amplitude ratio (AR): attenuation of amplitude, – Phase angle ( ): phase shift compared to input – These two quantities are the function of frequency.
Process
Input Output
sin
A ωt Aˆsin(ω φt+ )
After all transient effects are decayed out.
t φ
( ) u t ( )
y t
-A Aˆ A
Aˆ
−
φ
ˆ/
A A
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BENEFITS OF FREQUENCY RESPONSE
• Frequency responses are the informative representations of dynamic systems
– Audio Speaker
– Equalizer
– Structure
Elec. Signal Sound wave
Raw signal
Processed signal
Expensive speaker
Cheap speaker
Old Tacoma bridge New Tacoma bridge
Old Tacoma bridge
New Tacoma bridge Adjustable
for each frequency
band
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– Low-pass filter
– High-pass filter
– In signal processing field, transfer functions are called“filters ”.
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• Any linear dynamical system is completely defined by its frequency response.
– The AR and phase angle define the system completely.
– Bode diagram
• AR in log -log plot
• Phase angle in log -linear plot
– Via efficient numerical technique (fast Fourier transform, FFT), the output can be calculated for any type of input.
• Frequency response representation of a system dynamics is very convenient for designing a feedback controller and analyzing a closed-loop system.
– Bode stability
– Gain margin (GM) and phase margin (PM)
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• Critical frequency
– As controller gain changes, the amplitude ratio (AR) and the phase angle (PA) change.
– The frequency where the PA reaches –180° is calledcritical frequency ( ).
– The component of output at the critical frequency will have the exactly same phase as the signal goes through the loop due to comparator (-180 °) and phase shift of the process (-180 °).
– For the open-loop gain at the critical frequency,
• No change in magnitude
• Continuous cycling – For
• Getting bigger in magnitude
• Unstable – For
• Getting smaller in magnitude
• Stable ωc
( ) 1
OL c
K ω =
R Gc… GP
+
-
Y E
Sign change
Sign change
( ) 1
OL c
K ω >
( ) 1
OL c
K ω <
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• Example
– If a feed is pumped by a peristaltic pump to a CSTR, will the fluctuation of the feed flow appear in the output?
– V=50cm3, q=90cm3/min (so is the average of qi)
• Process time constant=0.555min.
– The rpm of the peristaltic pump is 60rpm.
• Input frequency=180rad/min (3blades) – The AR=0.01
t qi
V cAi, qi
Peristaltic pump cA, q
If the magnitude of fluctuation of qiis 5% of nominal flow rate, the fluctuation in the output concentration will be about 0.05% which is almost unnoticeable.
( constant)
A
i Ai A
Vdc q c qc q
dt = − ≈
/ ( )
( ) ( / ) 1
Ai A i
A i
C C q
C s
q s =Vs q= V q s
+ +
(ωτ =100)
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OBTAINING FREQUENCY RESPONSE
• From the transfer function, replace s with
– For a pole, , the response mode is .
– If the modes are not unstable ( ) and enough time elapses, the survived modes becomes . (ultimate response)
• The frequency response, is complex as a function of frequency.
jω
( )
s j( ) G s
= ω→ G j ω
Transfer function Frequency response
( ) G jω ( ) Re[ ( )] Im[ ( )]
G jω = G jω + j G jω
2 2
( ) Re[ ( )] Im[ ( )]
AR= G jω = G jω + G jω
( )
( ) tan 1 Im[ ( )]/Re[ ( )]
G j G j G j
φ=R ω = − ω ω
s= +α jω e(α+jω)t α≤0
ej tω
Re Im
φ ( ) G jω
AR
ω↑
Nyquist diagram
Bode plot
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• Getting ultimate response
– For a sinusoidal forcing function – Assume G(s) has stable poles bi.
– Without calculating transient response, the frequency response can be obtained directly from .
– Unstable transfer function does not have a frequency response because a sinusoidal input produces an unstable output response.
2 2
( ) ( ) A Y s G s
s ω
= ω
+
1
2 2 2 2
1
( ) ( ) n
n
A Cs D
Y s G s
s s b s b s
α
ω α ω
ω ω
= = + + + +
+ + L + +
( ) ( ) D C
G j A Cj D G j j R jI
A A
ω ω= ω+ ω⇒ ω = + = +
Decayed out at larget
, ( cosul sin ) ˆsin( )
C=IA D=RA⇒ y =A I ωt+R ωt = A ωt+φ
2 2 1
ˆ / = ( ) and tan ( / ) ( )
AR A A R I G jω φ − I R G jω
∴ = = + = =R
( )
G jω
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• First-order process
• Second-order process
( ) ( 1) G s K
τs
= +
2 2
( ) (1 )
(1 ) (1 )
K K
G j j
ω j ωτ
ωτ ω τ
= = −
+ +
2 2
( ) 1
N 1 AR G jω
= = ω τ + ( ) tan (1 ) φ=RG jω = − − ωτ
( ) 2 2
( 2 1)
G s K
s s
τ ζτ
= + +
2 2
( )
(1 ) 2
G j K ω j
τ ω ζτω
= − +
2 2 2 2
( )
(1 ) ( 2 )
A R G jω K
ω τ ζ ω τ
= =
− +
1 1
2 2
I m ( ( )) 2
( ) tan t a n
R e ( ( )) 1
G j G j
G j
ω ζωτ
φ ω
ω ω τ
− −
= = = −
R −
ζ>1
Corner freq.
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• Process Zero (lead)
• Unstable pole
( ) a 1 G s =τ s+
( ) 1 a
G jω = +jωτ
2 2
( ) 1
N a
AR =G jω = +ω τ ( ) tan (1 a) φ=RG jω = − ωτ
( ) 1
( 1)
G s = τs
− +
2 2
1 1
( ) (1 )
1 1
G j j
ω j τω
τω τ ω
= = +
− +
2 2
( ) 1 1 A R G jω
= = ω τ +
1I m ( ( )) 1
( ) tan tan
R e ( ( ) ) G j G j
G j
φ ω ω ωτ
ω
− −
=R = =
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• Integrating process
• Differentiator
• Pure delay process
( ) 1
G s =As G j( ) 1 1 j
jA A
ω = ω= − ω ( ) 1
ARN G j
ω Aω
= =
1 1
( ) tan ( )
0 2
G j π
φ ω
ω
= = − − = −
R ⋅
( ) s G s =e−θ
( ) j cos sin
G jω = e−θω= θω− j θω
( ) 1
A R= G jω =
( ) tan 1tan
φ =RG jω = − − θω = −θω ( )
G s =As G j( ω)= jAω ( )
ARN =G jω =Aω
1 1
( ) tan ( )
0 2
G j π
φ ω
ω
= = − =
R ⋅
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SKETCHING BODE PLOT
• Bode diagram
– AR vs. frequency in log-log plot – PA vs. frequency in semi-log plot – Useful for
• Analysis of the response characteristics
• Stability of the closed-loop systemonly for open-loop stable systems with phase angle curves exhibit a single critical frequency.
1 2 3
( ) ( ) ( ) ( ) ( ) ( ) ( )
a b c
G s G s G s G s =G s G s G s L
L 1 2 3
( ) ( ) ( )
( )
( ) ( ) ( )
a b c
G j G j G j G j G j G j G j
ω ω ω
ω = ω ω ω L
L
1 2 3
( ) ( ) ( )
( )
( ) ( ) ( )
a b c
G j G j G j G j G j G j G j
ω ω ω
ω = ω ω ω L
L
1 2 3
( ) ( ) ( ) ( )
( ) ( ) ( )
a b c
G j G j G j G j
G j G j G j
ω ω ω ω
ω ω ω
= + + +
− − − −
R R R R L
R R R L
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• Amplitude Ratio on log-log plot
– Start from steady-state gain at =0. If GOLincludes either integrator or differentiator it starts at or 0.
– Each first-order lag (lead) adds to the slope –1 (+1) starting at the corner frequency.
– Each integrator (differentiator) adds to the slope –1 (+1) starting at zero frequency.
– A delays does not contribute to the AR plot.
• Phase angle on semi-log plot
– Start from 0° or -180° at =0 depending on the sign of steady- state gain.
– Each first-order lag (lead) adds 0 ° to phase angle at =0, adds -90° (+90°) to phase angle at = , and adds -45° (+45°) to phase angle at corner frequency.
– Each integrator (differentiator) adds -90° (+90°) to the phase angle for all frequency.
– A delay adds – to phase angle depending on the frequency.
∞
∞
ω
ω ω
θω
ω
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Examples 1. 2.
G s( )=(10s+1)(5Ks+1)(s+1)5(0.5 1) 0.5
( ) (20 1)(4 1) s e s
G s s s
+ −
= + +
G1
G2 G3
G4
G5
G1 G2 G3
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3. PI: 5. PID:
4. PD:
( ) C 1 1
I
G s K τ s
= +
( ) C 1 1 D
I
G s K s
s τ τ
= + +
Slope=-1 Slope=+1
1/ at 0
Notch I D
ω = τ τ φ= °
1/ at 45
b I
ω = τ φ= − °
Slope=-1
( )
( ) C 1 D G s =K +τ s
Slope=+1
1/ at 45
b D
ω = τ φ= − °
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NYQUIST DIAGRAM
• Alternative representation of frequency response
• Polar plot of ( is implicit)
– Compact (one plot)
– Wider applicability of stability analysis than Bode plot
– High frequency characteristics will be shrunk near the origin.
• Inverse Nyquist diagram: polar plot of
– Combination of different transfer function components is not easy as with Nyquist diagram as with Bode plot.
Re Im
φ ( ) G jω
AR
ω↑
Nyquist diagram
( ) Re[ ( )] Im[ ( )]
G jω = G jω +j G jω ( )
G jω ω
1 /G j(ω)